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KonstantinChe [14]
3 years ago
9

HELP I HAVE A CLASS SOON

Mathematics
1 answer:
Margarita [4]3 years ago
4 0

Answer:

ΔTFS ≅ ΔRFS (sas) because:

+) FT = FR (given)

+) FS ≅ FS (in the picture)

+) ∠TFS = ∠RFS ( SF bisects ∠TFR)

Step-by-step explanation:

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Answer:

V = 2143.57 cm^3

Step-by-step explanation:

We want to find the volume of the sphere

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2 years ago
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Find all polar coordinates of point P where P = ordered pair 3 comma negative pi divided by 3 .
jek_recluse [69]

Answer:

The all polar coordinates of P are:

(3 , -π/3) , (3 , 5π/3) , (-3 , 2π/3) , (-3 , -4π/3)

Step-by-step explanation:

* Lets study the polar coordinates of a point

- In polar coordinates there is an infinite number of coordinates

 for a given point.

- The polar coordinates of a point (x , y) is (r , θ), where

  r = √ ( x2 + y2 )

  θ = tan-1 ( y / x )

# Ex: the following four points are all coordinates for the same point.

* (5 , π/3) = (5 , −5π/3) = (−5 , 4π/3) =(−5 , −2π/3)

- These four points only represent the coordinates of the point without  

  rotating more than once

- So the point (r,θ) can be represented by any of the following

  coordinate pairs  (r , θ + 2π n) and (−r , θ + (2n + 1) π), where n is

  any integer.

* Now lets solve the problem

∵ P = (3 , -π/3)

∵ (r , θ + 2πn)

∴ r = 3 an d Ф = -π/3

- let n = 1

∴ P = (3 , -π/3 + 2π)

∴ P = (3 , 5π/3)

∵ P =(3 , -π/3)

∵ P = (-r , θ + (2n + 1) π)

Let n = 0

∴ P = (-3 , -π/3 + (2×0 + 1) π)

∴ P = (-3 , -π/3 + (0 + 1) π)

∴ P = (-3 , -π/3 + π)

∴ P = (-3 , 2π/3)

∵ P =(3 , -π/3)

∵ P = (-r , θ + (2n + 1) π)

Let n = -1

∴ P = (-3 , -π/3 + (2(-1) + 1) π)

∴ P = (-3 , -π/3 + (-2 + 1) π)

∴ P = (-3 , -π/3 + -π) = (-3 , -4π/3)

∴ P = (-3 , -4π/3)

5 0
2 years ago
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